An object's velocity after t seconds is v ( t ) = 32 − 2 t  feet per second?

(a) How many seconds does it take for the object to come to a stop (velocity = 0)?

(b) How far does the car travel during that time?

(c) How many seconds does it take the car to travel half the distance in part (b)?

1 Answer

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  • Ash
    Lv 7
    1 month ago

    v(t) = 32 - 2t

    (a) When v=0

    0 = 32-2t

    2t=32

    t = 16 s

    (b) initial velocity v(0) = 32 - 2(0) = 32 m/s

    final velocity, v(16) = 0

    distance travelled, s = ½(u+v)t

    s = ½(32 +0)16 = 256 m

    (c) a = v'(t) = -2

    s = ut + ½at²

    (256/2) = 32t + ½(-2)t²

    128 = 32t - t²

    t² - 32t + 128 = 0

    t = {-(-32)±√[(-32)²-4(1)(128)]}/(2*1)

    t= 27 or 4.7.. Ignore 27s as it is larger than 16s and can't have half the distance

    so t = 4.7 s

     

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